Order of int32_t to uint64_t casting

Question

Does the C++ standard guarantee whether integer conversion that both widens and casts away the sign will sign-extend or zero-extend?

The quick test:

int32_t s = -1;
uint64_t u = s;

produces an 0xFFFFFFFFFFFFFFFF under Xcode, but is that a defined behavior in the first place?

Answer 1

When you do

uint64_t u = s;

[dcl.init]/17.9 applies which states:

the initial value of the object being initialized is the (possibly converted) value of the initializer expression. A standard conversion sequence ([conv]) will be used, if necessary, to convert the initializer expression to the cv-unqualified version of the destination type; no user-defined conversions are considered.

and if we look in [conv], under integral conversions, we have

Otherwise, the result is the unique value of the destination type that is congruent to the source integer modulo 2^N, where N is the width of the destination type.

So what you are guaranteed to have happen is that -1 becomes the largest number possible to represent, -2 is one less then that, -3 is one less then -2 and so on, basically it "wraps around".

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In fact,

unsigned_type some_name = -1;

Is the canonical way to create a variable with the maximum value for that unsigned integer type.